Schwarzschild: the first exact solution, and what its two singularities are

Idea. \(r_s\) is linear in \(M\), so one number generates the table; the density is the same number cubed, and its inverse-square dependence on \(M\) is the surprise worth remembering.

1. The radii. \(r_{s,\odot} = 2GM_\odot/c^2 = 2\times1.32712440018\times10^{20}/ 8.98755179\times10^{16} = 2953.25\) m.

• Sun: \(r_s = 2.953\) km, \(r_s/R_\odot = 2953.25/6.957\times10^{8} = 4.245\times10^{-6}\).
• Earth: \(r_s = 2GM_\oplus/c^2 = 8.870\) mm, \(r_s/R_\oplus = 1.391\times10^{-9}\).
• Neutron star: \(r_s = 1.4\times2953.25 = 4134.6\) m, \(r_s/R = 4134.6/12\,000 = 0.345\).
• White dwarf: \(r_s = 0.6\times2953.25 = 1772.0\) m, \(r_s/R = 1.969\times10^{-4}\).

None has a horizon: in every case \(r_s \lt R\), so the Schwarzschild radius lies inside the matter, where \(T_{\mu\nu}\ne0\) and the vacuum solution does not apply. The neutron star is the interesting entry — a third of the way there, which is why it is the laboratory for strong-field gravity.

2. The density. \[\bar\rho = \frac{M}{\frac43\pi r_s^3} = \frac{3M}{4\pi}\Big(\frac{c^2}{2GM}\Big)^3 = \frac{3c^6}{32\pi G^3M^2}.\] For \(M = 10M_\odot = 1.98841\times10^{31}\) kg: \(\bar\rho = 1.84\times10^{17}\ \mathrm{kg\,m^{-3}}\), comparable with nuclear matter, \(2.3\times10^{17}\). For \(4.3\times10^{6}M_\odot\): \(\bar\rho = 9.97\times10^{5}\ \mathrm{kg\,m^{-3}}\), about a thousand times water.

3. Water. Setting \(\bar\rho = 1000\ \mathrm{kg\,m^{-3}}\), \[M = \sqrt{\frac{3c^6}{32\pi G^3\times1000}} = 2.70\times10^{38}\ \text{kg} = 1.36\times10^{8}M_\odot .\] A hole of a hundred million solar masses — an ordinary quasar engine — is, by this bookkeeping, as thin as water.

4. It is a mass divided by a Euclidean volume computed from a radius that is not a radius: \(r_s\) labels a sphere by its area, the region inside it is not a ball of anything, and the matter that made the hole is not spread through it.

Check. Dimensions: \(c^6/(G^3M^2)\) is \(\mathrm{(m/s)^6}/(\mathrm{m^3kg^{-1}s^{-2}})^3\mathrm{kg^2} = \mathrm{kg\,m^{-3}}\) ✓. Limit: as \(M\to\infty\), \(\bar\rho\to0\), consistent with \(r_s\propto M\) growing faster than \(M^{1/3}\) ✓. Cross-check on part 1: \(r_s/R = 2GM/(Rc^2) = 2|\Phi|/c^2\), and unit 35 quoted \(|\Phi|/c^2 = 2.12\times10^{-6}\) at the Sun's surface — twice that is \(4.24\times10^{-6}\) ✓.

Idea. The Koszul formula is mechanical; the point of the exercise is the second, independent check, which catches the errors the first one hides.

1. Four symbols. With \(g^{tt} = -1/(Ac^2)\), \(g^{rr} = 1/B\), \(g^{\varphi\varphi} = 1/(r^2\sin^2\theta)\):

• \(\Gamma^t_{tr} = \tfrac12g^{tt}\big(\partial_tg_{tr} + \partial_rg_{tt} - \partial_tg_{tr}\big) = \tfrac12\cdot\frac{-1}{Ac^2}\cdot(-A'c^2) = \frac{A'}{2A}\) ✓
• \(\Gamma^r_{rr} = \tfrac12g^{rr}\partial_rg_{rr} = \frac{B'}{2B}\) ✓
• \(\Gamma^r_{\varphi\varphi} = \tfrac12g^{rr}\big(2\partial_\varphi g_{r\varphi} - \partial_rg_{\varphi\varphi}\big) = -\frac{1}{2B}\partial_r(r^2\sin^2\theta) = -\frac{r\sin^2\theta}{B}\) ✓
• \(\Gamma^\varphi_{\theta\varphi} = \tfrac12g^{\varphi\varphi}\partial_\theta g_{\varphi\varphi} = \frac{1}{2r^2\sin^2\theta}\cdot2r^2\sin\theta\cos\theta = \cot\theta\) ✓

2. Three distinct indices. Let \(g\) be diagonal. Then \(g^{\mu\nu}\) is diagonal too, so the sum over \(\nu\) gives \(\nu = \mu\) only: \[\Gamma^\mu_{\alpha\beta} = \tfrac12g^{\mu\mu}\big(\partial_\alpha g_{\mu\beta} + \partial_\beta g_{\mu\alpha} - \partial_\mu g_{\alpha\beta}\big)\quad(\text{no sum}).\] If \(\mu, \alpha, \beta\) are distinct, all three components \(g_{\mu\beta}\), \(g_{\mu\alpha}\), \(g_{\alpha\beta}\) are off-diagonal, hence identically zero, so all three derivatives vanish. The condition needed is only that \(g\) be diagonal in a neighbourhood — so that the derivatives of the vanishing components also vanish — and nothing about which coordinates each component depends on. \(\square\)

3. Compatibility. \((r,\theta,\theta)\): the left side is \(\partial_rg_{\theta\theta} = 2r\). The right side is \(2\Gamma^l{}_{r\theta}g_{l\theta} = 2\Gamma^\theta{}_{r\theta}g_{\theta\theta} = 2\cdot\frac1r\cdot r^2 = 2r\) ✓.

\((r,t,t)\): left side \(\partial_rg_{tt} = -A'c^2\). Right side \(2\Gamma^l{}_{rt}g_{lt} = 2\Gamma^t{}_{rt}g_{tt} = 2\cdot\frac{A'}{2A}\cdot(-Ac^2) = -A'c^2\) ✓.

Check. Dimensions: \(\Gamma^t_{tr}\) is \(1/\)length, and it multiplies \(\dot r\dot t\) to give \(\ddot t\) ✓. Special case \(A = B = 1\): \(\Gamma^t_{tr} = 0\), \(\Gamma^r_{\varphi\varphi} = -r\sin^2\theta\), which are flat space's spherical-coordinate symbols ✓. Second special case: \(A = f\), \(B = 1/f\) reproduces unit 32's worked example 5 ✓.

Another way

Part 3 can be run as a test of the whole list at once: metric compatibility with the Levi-Civita connection is equivalent to \(\Gamma^\lambda_{\mu\nu}\) being given by the Koszul formula, so if all sixteen independent \((k,i,j)\) identities hold, every symbol is right. Two of them are enough paranoia in practice, and they should be chosen to involve different symbols.

Idea. Both components are four terms; the discipline is to list, for each term, exactly which index values survive before writing anything down.

1. \(R_{tt}\).

• Term 1: \(\partial_\mu\Gamma^\mu_{tt} = \partial_r\Gamma^r_{tt} = \frac{c^2A''}{2B} - \frac{c^2A'B'}{2B^2}\).
• Term 2: \(-\partial_t\Gamma^\mu_{\mu t} = 0\), since \(\Gamma^\mu_{\mu t} = \partial_t\ln\sqrt{-g} = 0\).
• Term 3: \(\Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{tt}\) has only \(\lambda = r\): \(\big(\frac{A'}{2A} + \frac{B'}{2B} + \frac2r\big)\frac{c^2A'}{2B}\).
• Term 4: \(-\Gamma^\mu_{t\lambda}\Gamma^\lambda_{\mu t}\) has \((\mu,\lambda) = (t,r)\) and \((r,t)\), each giving \(\Gamma^t_{tr}\Gamma^r_{tt}\): \(-2\cdot\frac{A'}{2A}\cdot\frac{c^2A'}{2B} = -\frac{c^2A'^2}{2AB}\).

Adding, and dividing out \(c^2\): \[\frac{A''}{2B} - \frac{A'B'}{2B^2} + \frac{A'^2}{4AB} + \frac{A'B'}{4B^2} + \frac{A'}{rB} - \frac{A'^2}{2AB} = \frac{A''}{2B} - \frac{A'^2}{4AB} - \frac{A'B'}{4B^2} + \frac{A'}{rB},\] which is the quoted expression, since \(\frac{A'}{4B}\big(\frac{A'}{A} + \frac{B'}{B}\big) = \frac{A'^2}{4AB} + \frac{A'B'}{4B^2}\) ✓.

2. \(R_{rr}\).

• Term 1: \(\partial_\mu\Gamma^\mu_{rr} = \partial_r\Gamma^r_{rr} = \partial_r\frac{B'}{2B} = \frac{B''}{2B} - \frac{B'^2}{2B^2}\).
• Term 2: \(-\partial_r\Gamma^\mu_{\mu r} = -\partial_r\big(\frac{A'}{2A} + \frac{B'}{2B} + \frac2r\big) = -\frac{A''}{2A} + \frac{A'^2}{2A^2} - \frac{B''}{2B} + \frac{B'^2}{2B^2} + \frac{2}{r^2}\).
• Term 3: \(\Gamma^\mu_{\mu r}\Gamma^r_{rr} = \big(\frac{A'}{2A} + \frac{B'}{2B} + \frac2r\big) \frac{B'}{2B}\).
• Term 4: \(-\Gamma^\mu_{r\lambda}\Gamma^\lambda_{\mu r}\). The pairs are \((t,t)\), \((r,r)\), \((\theta,\theta)\), \((\varphi,\varphi)\): \(-\big[\big(\frac{A'}{2A}\big)^2 + \big(\frac{B'}{2B}\big)^2 + \frac{1}{r^2} + \frac1{r^2}\big]\).

Terms 1 and 2 cancel the \(B''\) and the \(B'^2\) pieces; the \(2/r^2\) of term 2 cancels the \(-2/r^2\) of term 4; what is left is \[-\frac{A''}{2A} + \frac{A'^2}{2A^2} + \frac{A'B'}{4AB} + \frac{B'^2}{4B^2} + \frac{B'}{rB} - \frac{A'^2}{4A^2} - \frac{B'^2}{4B^2} = -\frac{A''}{2A} + \frac{A'^2}{4A^2} + \frac{A'B'}{4AB} + \frac{B'}{rB},\] which is the quoted expression ✓.

3. The combination. Multiplying \(R_{tt}\) by \(B/(c^2A)\) gives \(\frac{A''}{2A} - \frac{A'}{4A}\big(\frac{A'}{A} + \frac{B'}{B}\big) + \frac{A'}{rA}\). The first two terms are exactly minus the first two of \(R_{rr}\), so the sum is \(\frac{A'}{rA} + \frac{B'}{rB} = \frac{(AB)'}{rAB}\) ✓. Such a combination had to exist because \(R_{tt}\) and \(R_{rr}\) are both second order in \(A\) and the \(A''\) enters each of them through the same object, \(\Gamma^r_{tt}\) in one case and \(\Gamma^\mu_{\mu r}\) in the other, with opposite signs; more structurally, the contracted Bianchi identity guarantees one relation among the three equations, and this is its first-order shadow.

Check. Dimensions: \(R_{tt}\) carries \(c^2\) because \(g_{tt}\) does, and \(R_{rr}\) does not ✓. Flat: \(A = B = 1\) kills both ✓. Schwarzschild, \(A = f\), \(B = 1/f\): the brackets vanish, leaving \(R_{tt} = c^2f(\frac{f''}2 + \frac{f'}r) = 0\) and \(R_{rr} = -\frac{1}{f}(\frac{f''}{2} + \frac{f'}{r}) = 0\) ✓ — both by the same cancellation \(f'' /2 = -r_s/r^3 = -f'/r\).

Idea. Time dependence enters the Ricci tensor first through the one component that was previously forbidden by symmetry, \(R_{tr}\); that component is proportional to \(\dot B\), so vacuum kills the time dependence of \(B\) immediately, and \(A\)'s survives only as a factor that a change of clock removes.

1. The new symbols. From the Lagrangian \(\mathcal L = -Ac^2\dot t^2 + B\dot r^2 + \ldots\), the \(t\) equation is now \(\frac{\mathrm d}{\mathrm d\lambda}(-2Ac^2\dot t) = \partial_t\mathcal L = -\dot Ac^2\dot t^2 + \dot B\dot r^2\), i.e. \[\ddot t + \frac{\dot A}{2A}\dot t^2 + \frac{A'}{A}\dot t\dot r + \frac{\dot B}{2Ac^2}\dot r^2 = 0,\] giving \(\Gamma^t_{tt} = \dot A/2A\), \(\Gamma^t_{tr} = A'/2A\) (unchanged) and \(\Gamma^t_{rr} = \dot B/(2Ac^2)\). The \(r\) equation gains \(\frac{\mathrm d}{\mathrm d\lambda}(2B\dot r) \supset 2\dot B\dot t\dot r\), giving \(\Gamma^r_{tr} = \dot B/2B\); all the old symbols keep their form. And \(\Gamma^\mu_{\mu t} = \partial_t\ln\sqrt{-g} = \partial_t\ln(c\sqrt{AB}r^2\sin\theta) = \frac{\dot A}{2A} + \frac{\dot B}{2B}\).

2. \(R_{tr}\).

• Term 1: \(\partial_\mu\Gamma^\mu_{rt} = \partial_t\Gamma^t_{rt} + \partial_r\Gamma^r_{rt} = \partial_t\frac{A'}{2A} + \partial_r\frac{\dot B}{2B}\).
• Term 2: \(-\partial_r\Gamma^\mu_{\mu t} = -\partial_r\frac{\dot A}{2A} - \partial_r\frac{\dot B}{2B}\). Since \(\partial_t\frac{A'}{2A} = \partial_t\partial_r\tfrac12\ln A = \partial_r\frac{\dot A}{2A}\), terms 1 and 2 cancel entirely.
• Term 3: \(\Gamma^\mu_{\mu\lambda}\Gamma^\lambda_{rt}\) with \(\lambda = t\) and \(\lambda = r\): \(\big(\frac{\dot A}{2A} + \frac{\dot B}{2B}\big)\frac{A'}{2A} + \big(\frac{A'}{2A} + \frac{B'}{2B} + \frac2r\big)\frac{\dot B}{2B}\).
• Term 4: \(-\Gamma^\mu_{r\lambda}\Gamma^\lambda_{\mu t}\). The nonzero products are \((\mu,\lambda) = (t,t)\): \(\Gamma^t_{rt}\Gamma^t_{tt} = \frac{A'}{2A}\frac{\dot A}{2A}\); \((r,t)\): \(\Gamma^r_{rt}\Gamma^t_{rt} = \frac{\dot B}{2B}\frac{A'}{2A}\); \((t,r)\): \(\Gamma^t_{rr}\Gamma^r_{tt} = \frac{\dot B}{2Ac^2}\cdot\frac{c^2A'}{2B} = \frac{A'\dot B}{4AB}\); \((r,r)\): \(\Gamma^r_{rr}\Gamma^r_{rt} = \frac{B'}{2B}\frac{\dot B}{2B}\). All four enter with a minus sign.

Term 3 expands to \(\frac{\dot AA'}{4A^2} + \frac{\dot BA'}{4AB} + \frac{A'\dot B}{4AB} + \frac{B'\dot B}{4B^2} + \frac{\dot B}{rB}\), and term 4 subtracts \(\frac{\dot AA'}{4A^2} + \frac{\dot BA'}{4AB} + \frac{A'\dot B}{4AB} + \frac{B'\dot B}{4B^2}\). Everything cancels except \[R_{tr} = \frac{\dot B}{rB}. \qquad\square\]

3. The rest of the proof. Vacuum gives \(\dot B = 0\), so \(B = B(r)\). The component \(R_{\theta\theta}\) involves only \(\Gamma^r_{\theta\theta}\), \(\Gamma^\mu_{\mu r}\) and the angular symbols, none of them differentiated in \(t\), so it keeps its static form \(1 - \frac1B + \frac{r}{2B}\big(\frac{B'}{B} - \frac{A'}{A}\big)\). Setting it to zero, \[\frac{A'}{A} = \frac{B'}{B} + \frac{2(B-1)}{r},\] whose right-hand side is a function of \(r\) alone; integrating in \(r\) at fixed \(t\), \(\ln A = F(r) + h(t)\), i.e. \(A = e^{h(t)}a(r)\) with \(a = e^F\). Define a new time by \(\mathrm d\tilde t = e^{h(t)/2}\mathrm dt\); then \(A c^2\mathrm dt^2 = a(r)c^2\mathrm d\tilde t^2\) and the metric is the static ansatz, whose unique vacuum solution is Schwarzschild (deck, section 2). \(\square\)

4. Consequences. (i) No monopole radiation. A round star that pulsates or collapses has an exterior that is Schwarzschild with a constant \(r_s\) at every instant, so nothing propagates outwards. This uses the whole theorem, and in particular step 1: it is \(\dot B = 0\) that forbids a time-dependent exterior. (ii) A flat cavity. Inside an empty round cavity, the theorem gives Schwarzschild with some \(r_s\), and regularity at the centre — which would otherwise carry a curvature singularity, since \(K = 12r_s^2/r^6\) — forces \(r_s = 0\). This uses the theorem's conclusion plus the invariant of P9.

Check. Dimensions: \(\dot B/(rB)\) is \(\mathrm{s^{-1}m^{-1}}\), and \(R_{tr}\) should be \([g_{tr}]/\text{length}^2 = \mathrm{m\,s^{-1}}/\mathrm{m^2}\) ✓. Limit: \(B\) static gives \(R_{tr} = 0\), recovering the symmetry argument of the deck, which said \(R_{tr} = 0\) for a static metric ✓. Special case \(A = B = 1\): flat, all components zero ✓.

Idea. Rotations act only on the sphere factor, and the sphere factor is \(r^2\) times a metric they are already known to preserve; the check is then three short lines. The charges they produce are the three components of angular momentum, and their vanishing in two directions is what confines an orbit to a plane.

1. The three components. Write \(R_x^\theta = -\sin\varphi\), \(R_x^\varphi = -\cot\theta\cos\varphi\).

• \((\theta\theta)\): \(R_x^\lambda\partial_\lambda(r^2) + 2g_{\theta\theta}\partial_\theta R_x^\theta = 0 + 2r^2\partial_\theta(-\sin\varphi) = 0\) ✓.
• \((\varphi\varphi)\): \(R_x^\theta\partial_\theta(r^2\sin^2\theta) + 2g_{\varphi\varphi}\partial_\varphi R_x^\varphi = -\sin\varphi\cdot2r^2\sin\theta\cos\theta + 2r^2\sin^2\theta\cdot\cot\theta\sin\varphi = 0\) ✓.
• \((\theta\varphi)\): \(0 + g_{\varphi\varphi}\partial_\theta R_x^\varphi + g_{\theta\theta}\partial_\varphi R_x^\theta = r^2\sin^2\theta\cdot\frac{\cos\varphi}{\sin^2\theta} + r^2(-\cos\varphi) = 0\) ✓.

The components with a \(t\) or an \(r\) index vanish because \(R_x\) has no such components and \(g_{tt}, g_{rr}\) do not depend on \(\theta, \varphi\); for instance \((\mathcal L_{R_x}g)_{r\theta} = g_{\theta\theta}\partial_rR_x^\theta = 0\).

2. The charge. \[g(R_x,\dot x) = g_{\theta\theta}R_x^\theta\dot\theta + g_{\varphi\varphi}R_x^\varphi\dot\varphi = -r^2\big(\sin\varphi\,\dot\theta + \sin\theta\cos\theta\cos\varphi\,\dot\varphi\big).\] Similarly \(g(R_y,\dot x) = r^2(\cos\varphi\,\dot\theta - \sin\theta\cos\theta\sin\varphi\,\dot\varphi)\) and \(g(\partial_\varphi,\dot x) = r^2\sin^2\theta\,\dot\varphi = L\).

3. The plane. At the initial event, \(\theta = \pi/2\) makes \(\cos\theta = 0\) and \(\dot\theta = 0\), so both \(g(R_x,\dot x)\) and \(g(R_y,\dot x)\) vanish, and they vanish for ever. Suppose at some later time \(\theta\ne\pi/2\). Then \(\cos\theta \ne 0\), and the two conserved equations read \(\sin\varphi\,\dot\theta = -\sin\theta\cos\theta\cos\varphi\,\dot\varphi\) and \(\cos\varphi\,\dot\theta = \sin\theta\cos\theta\sin\varphi\,\dot\varphi\); multiplying the first by \(\sin\varphi\), the second by \(\cos\varphi\) and adding gives \(\dot\theta = 0\), and substituting back gives \(\sin\theta\cos\theta\,\dot\varphi = 0\), hence \(\dot\varphi = 0\) as well — so the particle moves radially, and a radial curve through a point with \(\theta = \pi/2\) has \(\theta = \pi/2\) throughout, a contradiction. The one-line argument: \(\theta\equiv\pi/2\) solves the \(\theta\) geodesic equation \(\ddot\theta + \frac2r\dot r\dot\theta - \sin\theta\cos\theta\,\dot\varphi^2 = 0\) with these initial data, and by Picard–Lindelöf (unit 01) solutions with given initial data are unique.

4. Inside. \(g(\partial_t,\partial_t) = -(1 - r_s/r)c^2\), which for \(r \lt r_s\) is \(+(r_s/r - 1)c^2 \gt 0\): spacelike. The quantity \(E = -g(\partial_t,\dot x)\) is still conserved — the proof used only the Killing property — but it is no longer an energy: it is the momentum conjugate to translation along a spatial direction, and it may take either sign.

Check. Dimensions: \(g(R_x,\dot x)\) is \(\mathrm{m^2s^{-1}}\), like \(L\) ✓. Special case \(\theta = \pi/2\), \(\varphi = 0\): \(g(R_x,\dot x) = 0\) and \(g(R_y,\dot x) = r^2\dot\theta\) — the charge that measures tipping out of the equatorial plane ✓. Flat limit \(r_s\to0\): the three charges are the three components of \(\vec r\times\vec v\) in spherical coordinates, whose constancy is unit 08's planar motion ✓.

Idea. In proper time and areal radius, radial free fall obeys exactly Newton's energy equation, so it is Newton's cycloid — and the whole relativistic content is in the other clock, whose reading is a logarithm that never ends.

1. Worked example 3 with \(L = 0\): \(\dot r^2 = E^2/c^2 - (1 - r_s/r)c^2\). At \(r = r_0\), \(\dot r = 0\), so \(E^2/c^2 = (1 - r_s/r_0)c^2\), that is \(E = c^2\sqrt{1 - r_s/r_0}\) (positive root, future-directed). Substituting, \[\dot r^2 = c^2\Big[\Big(1 - \frac{r_s}{r_0}\Big) - \Big(1 - \frac{r_s}{r}\Big)\Big] = c^2r_s\Big(\frac1r - \frac1{r_0}\Big).\] With \(r_0 = 10r_s\), \(E = c^2\sqrt{0.9} = 0.948683\,c^2\).

2. The cycloid. Put \(r = \frac{r_0}{2}(1 + \cos\eta) = r_0\cos^2(\eta/2)\). Then \(\frac1r - \frac1{r_0} = \frac{1}{r_0}\big(\sec^2\frac\eta2 - 1\big) = \frac{\tan^2(\eta/2)}{r_0}\), so \(|\dot r| = c\sqrt{r_s/r_0}\,|\tan(\eta/2)|\). Also \(\mathrm dr = -r_0\cos\frac\eta2\sin\frac\eta2\,\mathrm d\eta\). Hence \[\mathrm d\tau = \frac{|\mathrm dr|}{|\dot r|} = \frac{r_0\cos\frac\eta2\sin\frac\eta2} {c\sqrt{r_s/r_0}\,\tan\frac\eta2}\,\mathrm d\eta = \frac{r_0}{c}\sqrt{\frac{r_0}{r_s}} \cos^2\frac\eta2\,\mathrm d\eta = \frac{r_0}{2c}\sqrt{\frac{r_0}{r_s}}(1 + \cos\eta)\,\mathrm d\eta,\] which integrates from \(\eta = 0\) (at \(r = r_0\)) to the stated \(\tau(\eta)\) ✓.

With \(r_0 = 10r_s = 295\,325\) m, the prefactor is \(\frac{r_0}{2c}\sqrt{10} = 1.5576\times10^{-3}\) s.

• To the centre, \(\eta = \pi\): \(\tau = 1.5576\times10^{-3}\times\pi = 4.8933\times10^{-3}\) s \( = 4.893\) ms.
• To the horizon: \(\cos\eta = 2r_s/r_0 - 1 = -0.8\), \(\eta = 2.49809\) rad, and \(\tau = 1.5576\times10^{-3}(2.49809 + 0.6) = 4.8255\times10^{-3}\) s \(= 4.826\) ms.
• Difference: \(6.776\times10^{-5}\) s \(= 67.76\ \mu\)s inside the horizon.

3. Coordinate time. \(\frac{\mathrm dt}{\mathrm dr} = \frac{\dot t}{\dot r} = -\frac{E}{c^2f\,|\dot r|}\). Near \(r = r_s\), \(f \approx (r - r_s)/r_s\) and \(|\dot r| \to c\sqrt{r_s(1/r_s - 1/r_0)} = c\sqrt{0.9}\), so \[\frac{\mathrm dt}{\mathrm dr} \approx -\frac{c^2\sqrt{0.9}}{c^3\sqrt{0.9}}\cdot\frac{r_s}{r - r_s} = -\frac{r_s}{c}\frac{1}{r - r_s},\qquad t \approx -\frac{r_s}{c}\ln\Big(\frac{r}{r_s} - 1\Big) + \text{const}.\] The coefficient is \(r_s/c = 98.51\ \mu\)s, and it does not depend on \(r_0\): every factor of \(e\) closer to the horizon costs the same \(98.51\ \mu\)s. Integrating \(\mathrm dt/\mathrm dr\) exactly from \(r_0\) to \(1.01r_s\) gives \(t = 59.84\,r_s/c = 5.894\times10^{-3}\) s \(= 5.894\) ms, against \(4.826\) ms of proper time to the horizon itself.

4. The reconciliation. The event "the body crosses \(r = r_s\)" is a point of the spacetime; the Schwarzschild chart does not contain it, because that chart covers only \(r \gt r_s\) (and, separately, \(r \lt r_s\)). So the crossing has no \(t\)-value at all, and "\(t\to\infty\)" is the statement that a coordinate runs out, not that a process slows down. The ingoing Eddington–Finkelstein coordinate \(v = ct + r_*\) assigns the crossing a perfectly finite value, and in that chart the body's worldline passes through \(r = r_s\) with \(\mathrm dv/\mathrm d\tau\) finite and positive.

Check. Dimensions: the cycloid prefactor is \((\text{m}/(\mathrm{m\,s^{-1}}))\times \text{dimensionless} = \)s ✓. Newtonian limit: for \(r_0\ll\) nothing in particular — set \(c^2r_s = 2GM\) and the equation \(\dot r^2 = 2GM(1/r - 1/r_0)\) is Newton's exactly, whose solution is the classical cycloid with period \(2\pi\sqrt{r_0^3/(8GM)}\); our \(\tau(\pi) = \frac{\pi r_0}{2c}\sqrt{r_0/r_s} = \frac{\pi}{2}\sqrt{r_0^3/(2GM)}\), which is half that period ✓, as it must be for a fall rather than a round trip. Number check on part 2: the last stretch, \(67.76\ \mu\)s, is longer than the \(65.67\ \mu\)s the deck computed for a body that arrives from rest at infinity, which is right — arriving slower means taking longer ✓.

Idea. \(g_{rr}\to\infty\) at the horizon, but as \((r - r_s)^{-1}\), so its square root is integrable: the horizon is a finite distance away, which is what makes it a place rather than an infinity.

1. On the slice, \(\mathrm d\ell^2 = \frac{\mathrm dr^2}{1 - r_s/r}\), so \(\ell = \int_{r_1}^{r_2}(1 - r_s/r)^{-1/2}\mathrm dr = r_s\int_{x_1}^{x_2}(1 - 1/x)^{-1/2}\mathrm dx\). Differentiating the proposed antiderivative, \[F'(x) = \frac{2x - 1}{2\sqrt{x(x-1)}} + \frac{\frac{1}{2\sqrt x} + \frac{1}{2\sqrt{x-1}}} {\sqrt x + \sqrt{x - 1}} = \frac{2x-1}{2\sqrt{x(x-1)}} + \frac{1}{2\sqrt{x(x-1)}} = \frac{x}{\sqrt{x(x-1)}} = \sqrt{\frac{x}{x-1}},\] which is \((1 - 1/x)^{-1/2}\) ✓. (The logarithm's derivative simplifies because \(\frac{1}{2\sqrt x} + \frac{1}{2\sqrt{x-1}} = \frac{\sqrt{x-1} + \sqrt x}{2\sqrt{x}\sqrt{x-1}}\).)

2. The numbers. \(F(1) = 0\), \(F(2) = \sqrt2 + \ln(1+\sqrt2) = 2.29559\), \(F(3) = \sqrt6 + \ln(\sqrt3+\sqrt2) = 2.44949 + 1.14622 = 3.59571\), \(F(4) = \sqrt{12} + \ln(2 + \sqrt3) = 3.46410 + 1.31696 = 4.78106\), \(F(10) = \sqrt{90} + \ln(\sqrt{10}+3) = 9.48683 + 1.81845 = 11.30528\), \(F(11) = \sqrt{110} + \ln(\sqrt{11}+\sqrt{10}) = 10.48809 + 1.86855 = 12.35664\).

• \(r_s\to2r_s\): \(\ell = 2.2956\,r_s\), against \(\Delta r = r_s\). For the hole, \(2.2956\times29.5325 = 67.79\) km.
• \(3r_s\to4r_s\): \(\ell = 1.1854\,r_s\).
• \(10r_s\to11r_s\): \(\ell = 1.0514\,r_s\).

3. Far away. \((1 - r_s/r)^{-1/2} = 1 + \frac{r_s}{2r} + O(r_s^2/r^2)\), so \(\ell \approx (r_2 - r_1) + \frac{r_s}{2}\ln\frac{r_2}{r_1}\). For \(10\to11\): \(\frac12\ln1.1 = 0.0477\), against the exact excess \(1.0514 - 1 = 0.0514\) — agreement to eight per cent, the discrepancy being the \(O(r_s^2/r^2)\) term ✓.

4. An integrand that diverges like \((x - 1)^{-1/2}\) has a finite integral, and the finite answer is what makes the horizon a surface at a definite distance below a static observer: one could, in principle, lower a rope to it. Had the integral diverged, the horizon would have been infinitely far away — as it is in the tortoise coordinate \(r_*\), which is a different statement, about light-travel time rather than about distance.

Check. Dimensions: \(\ell\) is \(r_s\) times a pure number ✓. Limit \(r_s\to0\): \(F(x)\to x\) for large \(x\) and \(\ell\to r_2 - r_1\), flat space ✓. Monotonicity: the three ratios \(2.296, 1.185, 1.051\) decrease toward 1 as the region moves outward, as the weakening field requires ✓.

Idea. A boost in the \((\hat t,\hat r)\) plane has unit determinant, and the Riemann tensor is antisymmetric in each pair of slots, so the boost leaves each pair alone; and the transverse components survive because the two terms a boost mixes are equal and opposite. Once that is known, the tidal formula is Newton's and the rest is arithmetic.

1. The two invariances. Write \(C = \cosh\chi\), \(S = \sinh\chi\).

Radial. Expand the first pair by multilinearity (unit 16) and use antisymmetry in the first two slots, which kills \(R(e_{\hat t}, e_{\hat t},\cdot,\cdot)\) and \(R(e_{\hat r}, e_{\hat r},\cdot, \cdot)\) and turns \(R(e_{\hat r}, e_{\hat t},\cdot,\cdot)\) into \(-R(e_{\hat t}, e_{\hat r},\cdot, \cdot)\): \[R(e_{\hat t}', e_{\hat r}',\cdot,\cdot) = (C^2 - S^2)\,R(e_{\hat t}, e_{\hat r},\cdot,\cdot) = R(e_{\hat t}, e_{\hat r},\cdot,\cdot),\] since \(C^2 - S^2 = 1\). The last two slots are the same pair, so the same reduction applies to them, and \(R_{\hat t'\hat r'\hat t'\hat r'} = R_{\hat t\hat r\hat t\hat r}\) ✓. (This is unit 33's remark that \(R\) sees each pair of slots only through the plane it spans, and a boost of unit determinant does not change that plane.)

Transverse. \(R(e_{\hat t}', e_{\hat\theta}, e_{\hat t}', e_{\hat\theta})\) expands, by multilinearity, into \[C^2R_{\hat t\hat\theta\hat t\hat\theta} + 2CS\,R_{\hat t\hat\theta\hat r\hat\theta} + S^2R_{\hat r\hat\theta\hat r\hat\theta}.\] The middle component vanishes (it is not on the list), and \(R_{\hat r\hat\theta\hat r\hat\theta} = -R_{\hat t\hat\theta\hat t\hat\theta}\), so the expression is \((C^2 - S^2)R_{\hat t\hat\theta\hat t\hat\theta} = R_{\hat t\hat\theta\hat t\hat\theta}\) ✓. \(\square\)

So an observer falling radially past radius \(r\), however fast, measures the same tidal tensor as a static observer would at that \(r\) — and since the components are finite at \(r = r_s\), where no static observer exists, the formula extends there by continuity.

2. The stretch. From the deviation equation, \(\ddot\xi^{\hat r} = \frac{2GM}{r^3}\xi^{\hat r}\), so across a body of proper length \(\ell\) the difference in acceleration between head and feet is \[\Delta a = \frac{2GM\ell}{r^3}, \qquad\text{at } r = r_s:\quad \Delta a = \frac{2GM\ell}{(2GM/c^2)^3} = \frac{c^6\ell}{4G^2M^2}.\]

3. The threshold. Require \(\frac{c^6\ell}{4G^2M^2} \le a_{\max} = 10g = 98.0665\ \mathrm{m\,s^{-2}}\): \[M \ge \frac{c^3}{2G}\sqrt{\frac{\ell}{a_{\max}}} = \frac{(2.99792458\times10^{8})^3} {2\times6.67430\times10^{-11}}\sqrt{\frac{2}{98.0665}} = 2.884\times10^{34}\ \text{kg} = 1.45\times10^{4}\,M_\odot .\] Above about fifteen thousand solar masses, a person crosses the horizon intact — and whether they survive is then a question about what happens later, not at the crossing.

4. Two holes. \(10\,M_\odot\): \(\Delta a = 2\times1.32712\times10^{21}\times2/(29\,532.5)^3 = 2.06\times10^{8}\ \mathrm{m\,s^{-2}} = 2.1\times10^{7}g\) — lethal by seven orders of magnitude. \(4.3\times10^{6}M_\odot\): \(r_s = 1.2699\times10^{10}\) m and \(\Delta a = 1.11\times10^{-3}\ \mathrm{m\,s^{-2}} = 1.1\times10^{-4}g\) — imperceptible.

Check. Dimensions: \(c^6\ell/(G^2M^2)\) is \(\mathrm{m^6s^{-6}\cdot m}/(\mathrm{m^6kg^{-2}s^{-4}}\cdot\mathrm{kg^2}) = \mathrm{m\,s^{-2}}\) ✓. Scaling: \(\Delta a\propto M^{-2}\), so between \(10\) and \(4.3\times10^{6}\) solar masses it should fall by \((4.3\times10^{5})^2 = 1.8\times10^{11}\); and \(2.06\times10^{8}/1.11\times10^{-3} = 1.9\times10^{11}\) ✓. Sanity: the Earth's tide across a two-metre person at its surface is \(2GM_\oplus\ell/R_\oplus^3 = 6.1\times10^{-6}\ \mathrm{m\,s^{-2}}\), so the supermassive hole's horizon is about 180 Earth-surface tides — noticeable to an instrument, not to a person ✓.

Idea. The contraction is a sum of squares because each nonzero component carries the time index twice or not at all; the arithmetic is then a count, and the physics is in how violently the answer falls with \(M\).

1. The contraction. In an orthonormal frame, \(R^{\alpha\beta\gamma\delta} = \eta^{\alpha\alpha'}\eta^{\beta\beta'}\eta^{\gamma\gamma'} \eta^{\delta\delta'}R_{\alpha'\beta'\gamma'\delta'}\), and each raised \(\hat t\) contributes \(-1\). Every nonzero component of Schwarzschild has either no \(\hat t\) index or exactly two, so each product \(R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}\) (no sum) equals \(+(R_{\alpha\beta\gamma\delta})^2\). The nonzero components are the six "sectional" ones \(R_{abab}\), \(a\ne b\), and for each unordered pair the symmetries \(R_{abab} = R_{baba} = -R_{abba} = -R_{baab}\) give four index orderings with the same square. Hence \[K = 4\sum_{\{a,b\}}(R_{abab})^2 = 4\Big[2\Big(\frac{r_s}{r^3}\Big)^2 + 4\Big(\frac{r_s}{2r^3}\Big)^2\Big] = 4\cdot3\,\frac{r_s^2}{r^6} = \frac{12r_s^2}{r^6} = \frac{48G^2M^2}{c^4r^6}.\]

2. Two numbers. At a horizon, \(K = 12/r_s^4\). For \(r_s = 29\,532.5\) m: \(K = 12/(29\,532.5)^4 = 1.58\times10^{-17}\ \mathrm{m^{-4}}\). At the Earth's surface, \(r_s = 8.870\times10^{-3}\) m and \(r = R_\oplus = 6.378137\times10^{6}\) m: \(K = 12\times(8.870\times10^{-3})^2/(6.378137\times10^{6})^6 = 1.40\times10^{-44}\ \mathrm{m^{-4}}\).

3. The crossover. \(12/r_s^4 = 1.402\times10^{-44}\) gives \(r_s = (12/1.402\times10^{-44})^{1/4} = 1.710\times10^{11}\) m, and \(M = r_sc^2/2G = 1.15\times10^{38}\) kg \(= 5.8\times10^{7}M_\odot\). A hole of some fifty-eight million solar masses has a horizon at which spacetime is less curved than it is in this room.

4. The logic. The colleague's argument is invalid: finiteness of one invariant, or even of all polynomial curvature invariants, does not imply that the geometry extends. Exact plane gravitational waves have every such invariant zero and are not flat; Minkowski space with a point deleted has all invariants zero and is singular by any reasonable definition. The correct argument for the horizon is constructive: exhibit a chart — ingoing Eddington–Finkelstein — in which the metric components are smooth and \(\det g = -r^4\sin^2\theta \ne 0\) at \(r = r_s\). That proves the geometry extends, because it displays the extension.

At \(r = 0\) the argument runs the other way and is immune to the objection because it is a proof by contradiction about every chart at once: \(K\) is a scalar, so its value along a curve does not depend on coordinates; a timelike geodesic reaches \(r\to0\) in finite proper time, with \(K = 12r_s^2/r^6\to\infty\) along it; if the metric extended as a \(C^2\) tensor field to a manifold containing that curve's endpoint, \(K\) would be continuous, hence bounded, near the endpoint. Contradiction. No chart can exist, so none need be searched for.

Check. Dimensions: \([K] = \mathrm{m^{-4}}\), the square of a curvature ✓. Limits: \(M\to0\) gives \(K\to0\); \(r\to\infty\) gives \(K\to0\) ✓. Scaling check on part 3: \(K\) at a horizon scales as \(M^{-4}\), so going from \(10\,M_\odot\) (\(1.58\times10^{-17}\)) down to \(1.40\times10^{-44}\) needs a factor \((1.13\times10^{27})^{1/4} = 5.8\times10^{6}\) in mass, i.e. \(5.8\times10^{7}M_\odot\) ✓.

Idea. Labelling events by the outgoing light ray that leaves them, rather than the ingoing one that arrives, produces an equally good chart across an equally good horizon — but the other one. A chart cannot be neutral about the direction in which it extends spacetime.

1. The metric. With \(u = ct - r_*\), \(c\,\mathrm dt = \mathrm du + \mathrm dr/f\), so \[-fc^2\mathrm dt^2 + \frac{\mathrm dr^2}{f} = -f\Big(\mathrm du + \frac{\mathrm dr}{f}\Big)^2 + \frac{\mathrm dr^2}{f} = -f\,\mathrm du^2 - 2\,\mathrm du\,\mathrm dr .\] The \((u,r)\) block is \(\begin{pmatrix}-f & -1\\ -1 & 0\end{pmatrix}\), determinant \(-1\) for every \(r\); with the angular part, \(\det g = -r^4\sin^2\theta\ne0\), and every component is smooth for \(r \gt 0\). So this chart too extends the exterior through \(r = r_s\).

2. Which way \(r\) goes. \(g_{rr} = 0\), so \(\partial_r\) is null; and \(g(c\partial_u, \partial_r) = c\,g_{ur} = -c \lt 0\), where \(c\partial_u = \partial_t\) at fixed \(r\) is future-directed outside. So \(+\partial_r\) is future-directed, and any future causal \(X\) has \(g(X, \partial_r) \le 0\), i.e. \(-X^u \le 0\): \(X^u \ge 0\). Now \[g(X,X) = -f(X^u)^2 - 2X^uX^r + r^2|X_\Omega|^2 \le 0 \ \Rightarrow\ 2X^uX^r \ge -f(X^u)^2 + r^2|X_\Omega|^2 \ge -f(X^u)^2 \gt 0\] for \(r \lt r_s\), where \(f \lt 0\); with \(X^u \gt 0\) this gives \(X^r \gt 0\). If \(X^u = 0\) then \(g(X,X) = r^2|X_\Omega|^2 \le 0\) forces \(X\propto\partial_r\), again with \(X^r \gt 0\). So \(r\) increases: nothing can stay inside, and nothing can enter.

3. Which regions. The ingoing chart \((v,r)\) covers region I (the exterior) together with region II (the black hole) and the future horizon between them. The outgoing chart \((u,r)\) covers region I together with region IV (the white hole) and the past horizon. Both contain the same exterior; they disagree about what else exists, because each one is built from light rays travelling one way.

4. Yes — both are honest extensions of the exterior, and the Kruskal chart of worked example 5 contains both at once, which is what "maximal" means.

Check. Dimensions: \(u\) is a length, \(\mathrm du\,\mathrm dr\) an area, as the other terms are ✓. Flat limit \(r_s\to0\): \(u = ct - r\) and \(-\mathrm du^2 - 2\mathrm du\,\mathrm dr = -c^2\mathrm dt^2 + \mathrm dr^2\) ✓. Consistency with part 2 of the deck's horizon theorem: the two results are time reverses of each other, and \(t\mapsto-t\) indeed exchanges \(u\) and \(v\) ✓.

Idea. Squash the two null coordinates separately with a bounded increasing function. Light cones do not notice, because a conformal factor cannot change the sign of \(g(X,X)\); so the picture is exact about causality and wrong about lengths, which is the trade one wants.

1. The metric. First, \(-c^2\mathrm dt^2 + \mathrm dr^2 = -\mathrm du\,\mathrm dv\) since \(u = ct - r\), \(v = ct + r\) give \(\mathrm du\,\mathrm dv = c^2\mathrm dt^2 - \mathrm dr^2\). Then \[-\mathrm du\,\mathrm dv = -\ell^2\sec^2q\sec^2p\,\mathrm dq\,\mathrm dp = -\frac{\ell^2}{\cos^2p\cos^2q}\,\mathrm dp\,\mathrm dq = \frac{\ell^2}{4\cos^2p\cos^2q}\big(-\mathrm dT^2 + \mathrm dR^2\big),\] using \(\mathrm dT^2 - \mathrm dR^2 = (\mathrm dp + \mathrm dq)^2 - (\mathrm dp - \mathrm dq)^2 = 4\,\mathrm dp\,\mathrm dq\). The angular part is \(r^2\mathrm d\Omega^2\), and part 2 gives \(r^2 = \frac{\ell^2\sin^2R}{4\cos^2p\cos^2q}\), the same prefactor times \(\sin^2R\). Range: \(u, v\) run over \(\mathbb R\) with \(v \ge u\), so \(p, q\in(-\pi/2,\pi/2)\) with \(p \ge q\), that is \(R \ge 0\) and \(|T| + R \lt \pi\).

2. The two formulas. \(r = \frac{v - u}{2} = \frac{\ell}{2}(\tan p - \tan q) = \frac{\ell}{2}\cdot\frac{\sin(p - q)}{\cos p\cos q} = \frac{\ell\sin R}{2\cos p\cos q}\), and \(2\cos p\cos q = \cos(p-q) + \cos(p+q) = \cos R + \cos T\), giving \(r/\ell = \frac{\sin R}{\cos R + \cos T}\) ✓. Identically, \(ct = \frac{u + v}{2} = \frac{\ell}{2}(\tan p + \tan q) = \frac{\ell\sin(p+q)}{2\cos p\cos q} = \frac{\ell\sin T}{\cos R + \cos T}\) ✓.

3. The boundary.

• \(i^+\): \(T = \pi\), \(R = 0\) (\(p = q = \pi/2\)), reached as \(t\to+\infty\) at bounded \(r\). Every timelike geodesic ends there; \(i^-\), at \(T = -\pi\), is where they start.
• \(\mathscr I^+\): the open segment \(T + R = \pi\), \(0 \lt R \lt \pi\) (\(p = \pi/2\), \(q\) finite) — where outgoing light rays end. \(\mathscr I^-\), \(T - R = -\pi\), is where ingoing rays begin.
• \(i^0\): \(T = 0\), \(R = \pi\), reached as \(r\to\infty\) at fixed \(t\). Every spacelike slice of constant \(t\) ends there.

4. Three curves. \(r = \) const is a timelike curve running from \(i^-\) to \(i^+\), bulging to the right (larger \(R\)) as \(r\) grows, since by part 2 a constant \(r\) requires \(\sin R \propto \cos R + \cos T\), which is largest at \(T = 0\). \(t = \) const runs from the left edge \(R = 0\) (which is \(r = 0\), the centre) to \(i^0\), passing through \(T = 0\) when \(t = 0\). A radial light ray has \(p\) constant on the way in — a line of slope \(-1\) in the \((R,T)\) plane — reaches the left edge \(R = 0\), and leaves with \(q\) constant, a line of slope \(+1\), ending on \(\mathscr I^+\). It is a "V" with its point on the axis \(r = 0\).

Check. Dimensions: \(\ell\) is the only length in the conformal factor; \(T, R\) are angles ✓. Special case \(T = R = 0\) (the origin of spacetime): the factor is \(\ell^2/4\) and \(r = 0\) ✓. Limit: for \(|u|, |v| \ll \ell\), \(p\approx v/\ell\), \(q \approx u/\ell\), so \(T \approx 2ct/\ell\), \(R \approx 2r/\ell\) and the metric reduces to \(\frac{\ell^2}{4}(-\mathrm dT^2 + \mathrm dR^2 + R^2\mathrm d\Omega^2) = -c^2\mathrm dt^2 + \mathrm dr^2 + r^2\mathrm d\Omega^2\) ✓ — near the origin the diagram is faithful.

Idea. The horizon sits at \(u = +\infty\) or \(v = -\infty\), and the exponential map sends those to finite values \(U = 0\), \(V = 0\). The factor \(f\), which vanishes there, is exactly cancelled by the Jacobian of the exponential, which blows up there.

1. The substitution. In region I, \(\mathrm ds^2 = -f\,\mathrm du\,\mathrm dv + r^2\mathrm d\Omega^2\) (this is the deck's step 2). Then \(\mathrm dU\,\mathrm dV = -\frac{UV}{4r_s^2}\,\mathrm du\,\mathrm dv\), so \(\mathrm du\,\mathrm dv = -\frac{4r_s^2}{UV}\,\mathrm dU\,\mathrm dV\). And \[UV = -e^{-u/2r_s}e^{v/2r_s} = -e^{(v - u)/2r_s} = -e^{r_*/r_s} = -e^{r/r_s}\Big(\frac{r}{r_s} - 1\Big) = \Big(1 - \frac{r}{r_s}\Big)e^{r/r_s}.\] Hence \[-f\,\mathrm du\,\mathrm dv = -\frac{r - r_s}{r}\cdot\Big(-\frac{4r_s^2}{UV}\Big)\mathrm dU\,\mathrm dV = -\frac{r - r_s}{r}\cdot\frac{4r_s^2}{(r/r_s - 1)e^{r/r_s}}\,\mathrm dU\,\mathrm dV = -\frac{4r_s^3}{r}e^{-r/r_s}\,\mathrm dU\,\mathrm dV,\] using \((r - r_s) = r_s(r/r_s - 1)\) ✓.

2. The level sets. \(UV = T^2 - X^2\), because \(T^2 - X^2 = \tfrac14[(V+U)^2 - (V-U)^2] = UV\). So \(r = \) const is \(T^2 - X^2 = (1 - r/r_s)e^{r/r_s}\), i.e. \(X^2 - T^2 = (r/r_s - 1)e^{r/r_s}\), a hyperbola: opening left and right for \(r \gt r_s\), up and down for \(r \lt r_s\) ✓. For \(t\): in region I, \(-\frac VU = e^{(u+v)/2r_s} = e^{ct/r_s}\), so \[\frac TX = \frac{V + U}{V - U} = \frac{-V/U - 1}{-V/U + 1} = \frac{e^{ct/r_s} - 1}{e^{ct/r_s} + 1} = \tanh\frac{ct}{2r_s},\] a straight line through the origin with slope in \((-1,1)\) ✓.

3. The Killing field. Under \(t\mapsto t + s\), \(u\mapsto u + cs\) and \(v\mapsto v + cs\), so \(U\mapsto e^{-cs/2r_s}U\), \(V\mapsto e^{cs/2r_s}V\). Differentiating at \(s = 0\), \[\partial_t = \frac{c}{2r_s}\big(V\partial_V - U\partial_U\big) = \frac{c}{2r_s}\big(X\partial_T + T\partial_X\big),\] the second form following from \(U = T - X\), \(V = T + X\). That is the generator of Lorentz boosts of the flat \((X, T)\) plane. Its norm: from the metric, \(g(\partial_t,\partial_t) = -\frac{4r_s^3}{r}e^{-r/r_s}\cdot\frac{c^2}{4r_s^2}\cdot 2\cdot \big(-UV\big)\cdot\frac12\), which is most simply evaluated from the Schwarzschild form as \(-(1 - r_s/r)c^2\): negative (timelike) for \(r \gt r_s\), zero on the horizon, positive (spacelike) for \(r \lt r_s\) ✓.

4. The bifurcation sphere. \(U = V = 0\) is the origin \(T = X = 0\), where the two horizons cross. There \(UV = 0\), so \((1 - r/r_s)e^{r/r_s} = 0\), giving \(r = r_s\): it is a sphere of area \(4\pi r_s^2\), the throat of the Einstein–Rosen bridge at \(T = 0\). It is the fixed-point set of the boost — it does not move under time translation, which is why the Killing field vanishes there. Its flat-spacetime analogue is unit 30's Rindler horizon and its bifurcation point at the origin of Minkowski space, where the two Rindler horizons \(X = \pm cT\) cross.

Check. Dimensions: \(U, V\) dimensionless, \(r_s^3/r\) an area ✓. Limit: far outside, \(|UV|\) is huge and \(r\) large; near the horizon \(UV\to0\) and \(r\to r_s\) ✓. The singularity: \(r\to0\) gives \(UV\to1\), the hyperbola \(T^2 - X^2 = 1\), which lies at \(|T| \ge 1\) — in the future of region II and the past of region IV, never in I or III ✓. A last consistency check: the Killing field is timelike in I, and \(T/X = \tanh(ct/2r_s)\) shows the \(t\)-lines fan out from the origin, filling exactly the wedge \(|T| \lt X\), which is region I ✓.

Idea. Inside the horizon \(r\) is a time, and every future-directed curve must spend it. The proper time elapsed is largest when nothing else is happening — no motion in \(t\), no motion in the angles — and that is a specific geodesic, not a hovering observer, because hovering is not available.

1. The inequality. For \(r \lt r_s\), write \(F := r_s/r - 1 \gt 0\). Then \(f = -F\) and \[\mathrm ds^2 = Fc^2\mathrm dt^2 - \frac{\mathrm dr^2}{F} + r^2\mathrm d\Omega^2 .\] The coefficient of \(\mathrm dr^2\) is negative and that of \(\mathrm dt^2\) positive: \(r\) is the timelike coordinate. Along a timelike curve, \(\mathrm ds^2 = -c^2\mathrm d\tau^2\), so \[c^2\mathrm d\tau^2 = \frac{\mathrm dr^2}{F} - Fc^2\mathrm dt^2 - r^2\mathrm d\Omega^2 \le \frac{\mathrm dr^2}{F},\] since the two subtracted terms are non-negative. Taking square roots, and noting that \(r\) decreases monotonically (the deck's horizon theorem), \(c\,\mathrm d\tau \le |\mathrm dr|/\sqrt{r_s/r - 1}\), and integrating from \(r = r_s\) down to \(r = 0\), \[c\,\tau \le \int_0^{r_s}\frac{\mathrm dr}{\sqrt{r_s/r - 1}} .\]

2. The integral. Substitute \(r = r_s\sin^2\psi\), \(\psi: 0\to\pi/2\). Then \(\mathrm dr = 2r_s\sin\psi\cos\psi\,\mathrm d\psi\) and \(\sqrt{r_s/r - 1} = \sqrt{\frac{1 - \sin^2\psi}{\sin^2\psi}} = \frac{\cos\psi}{\sin\psi}\), so the integrand becomes \(2r_s\sin^2\psi\,\mathrm d\psi\) and \[\int_0^{r_s}\frac{\mathrm dr}{\sqrt{r_s/r - 1}} = 2r_s\int_0^{\pi/2}\sin^2\psi\,\mathrm d\psi = 2r_s\cdot\frac\pi4 = \frac{\pi r_s}{2}.\] Hence \(\tau \le \frac{\pi r_s}{2c} = \frac{\pi GM}{c^3}\).

3. The extremal curve. Equality needs \(\mathrm d\Omega = 0\) and \(\mathrm dt = 0\) — no angular motion, and \(t\) constant. Take the radial geodesic with \(E = 0\): worked example 3 gives \(\dot r^2 = E^2/c^2 - fc^2 = -fc^2 = c^2(r_s/r - 1)\), which is exactly the equality case, and \(\dot t = E/(fc^2) = 0\), so \(t\) is indeed constant. It is the limit of falls released from rest at \(r_0\to r_s^+\): a body that is dropped from just outside the horizon, with no outward or sideways motion at all, lives longest.

4. The numbers, and the advice. \(10\,M_\odot\): \(\tau_{\max} = \pi\times29\,532.5/(2\times2.99792458\times10^{8}) = 1.547\times10^{-4}\) s \(= 154.7\ \mu\)s. \(4.3\times10^{6}M_\odot\): \(r_s = 1.2699\times10^{10}\) m and \(\tau_{\max} = 66.54\) s. The advice: do not fire the engines. Any thrust — outward, to resist the fall, or in any other direction — makes \(\mathrm dt\) or \(\mathrm d\Omega\) nonzero and strictly shortens the remaining proper time. Free fall, as quietly as possible, is the longest life available.

Check. Dimensions: \(GM/c^3\) is \(\mathrm{m^3s^{-2}}/\mathrm{m^3s^{-3}} = \) s ✓. Comparison: a body falling from rest at infinity has \(2r_s/3c = 65.7\ \mu\)s left at the horizon for the \(10\,M_\odot\) hole, less than the maximum \(154.7\ \mu\)s, as the bound demands, and the ratio is \(\frac{2/3}{\pi/2} = 0.424\) independently of \(M\) ✓. Limit: \(M\to0\) gives \(\tau_{\max}\to0\), no interior to have a life in ✓. Order of magnitude: \(\tau_{\max} = 1.57\,r_s/c\), so the longest possible life inside is about the time light takes to cross the hole — which is the honest summary of what a black hole is ✓.